The general solution of the equation
`"sin" 2x+ 2 "sin" x+ 2 "cos" x + 1 = 0`is

Options

A. `3n pi - (pi)/(4), n in Z`
B. `2n pi + (pi)/(4), n in Z`
C. `2 n pi + (-1)^(n) "sin"^(-1) ((1)/(sqrt(3))), n in Z`
D. `n pi - (pi)/(4), n in Z`

Explanation

To solve the equation \( \sin 2x + 2 \sin x + 2 \cos x + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation using the double angle formula
We know that \( \sin 2x = 2 \sin x \cos x \). Substituting this into the equation gives:
\[ 2 \sin x \cos x + 2 \sin x + 2 \cos x + 1 = 0 \]

### Step 2: Factor out common terms
We can factor out 2 from the first three terms:
\[ 2 (\sin x \cos x + \sin x + \cos x) + 1 = 0 \]
This simplifies to:
\[ 2 (\sin x \cos x + \sin x + \cos x) = -1 \]

### Step 3: Rearrange the equation
Now, we can isolate the expression in parentheses:
\[ \sin x \cos x + \sin x + \cos x = -\frac{1}{2} \]

### Step 4: Use the identity \( \sin^2 x + \cos^2 x = 1 \)
We can express \( \sin x + \cos x \) in terms of a single variable. Let \( y = \sin x + \cos x \). Then:
\[ \sin^2 x + \cos^2 x = 1 \implies y^2 = 1 + 2\sin x \cos x \]
Thus, \( \sin x \cos x = \frac{y^2 - 1}{2} \).

### Step 5: Substitute back into the equation
Substituting this back into our rearranged equation gives:
\[ \frac{y^2 - 1}{2} + y = -\frac{1}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ y^2 - 1 + 2y = -1 \]
This simplifies to:
\[ y^2 + 2y = 0 \]

### Step 6: Factor the quadratic equation
Factoring gives:
\[ y(y + 2) = 0 \]
Thus, we have two solutions:
\[ y = 0 \quad \text{or} \quad y + 2 = 0 \implies y = -2 \]

### Step 7: Solve for \( \sin x + \cos x = 0 \)
The equation \( \sin x + \cos x = 0 \) can be rewritten as:
\[ \sin x = -\cos x \]
This implies:
\[ \tan x = -1 \]
The general solution for this is:
\[ x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Step 8: Check the second case \( y = -2 \)
The equation \( \sin x + \cos x = -2 \) is not possible since the maximum value of \( \sin x + \cos x \) is \( \sqrt{2} \) and the minimum is \( -\sqrt{2} \), which means this case does not yield any solutions.

### Final Answer
Thus, the general solution of the equation \( \sin 2x + 2 \sin x + 2 \cos x + 1 = 0 \) is:
\[ x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

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